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How to evaluate the limit: $$\lim_{x \to 0} \Bigl(\frac{\sin{x}}{x}\Bigr)^{1/x^{3}}$$

I think it goes to $1$ because $\lim\limits_{x \to 0} \frac{\sin{x}}{x} =1$ and so power of $1$ should also be $1$. Am I right?

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No $1^{\infty}$ is considered to be an indeterminate form. –  user9413 Sep 3 '11 at 13:35
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For comparison, consider the standard limit $e = \lim_{x \to 0} (1+x)^{1/x}$. Applying your logic would give $1$ as the limit. (By the way, this limit could be useful for you depending on how you approach the problem...) –  Srivatsan Sep 3 '11 at 13:40
    
A question on related the $1^{\infty}$ indeterminate form: math.stackexchange.com/questions/10490/… –  Srivatsan Sep 3 '11 at 14:12
    
I think a more interesting problem would be $\lim_{x \to 0} \Bigl(\frac{\sin{x}}{x}\Bigr)^{1/x^{2}}$ –  robjohn Sep 4 '11 at 0:23

4 Answers 4

You are wrong. For example, $$\lim_{x\to 0}\left(1+x\right)^{\frac{1}{x}}=e$$ But by your approach it would be $1$. This is because $1^\infty$ is undefined and not $1$. The correct approach would be to use the limit I gave as example to get: $$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^3}}=\lim_{x\to 0}\left(1+\frac{\sin x-x}{x}\right)^{\frac{x}{\sin x-x}\frac{1}{x^3}\frac{\sin x-x}{x}}=\lim_{x\to 0}\,\,\,e\,\,^{\frac{\sin x-x}{x^4}}$$ You can continue from here.

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I sure hope the downvoter would care to explain him/herself... –  J. M. Sep 3 '11 at 15:07
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Well you guys know I always kiss and tell... my best guess is that someone is reading something into the statement "You are wrong". It's not like Dennis was demeaning in any way; he answered the question!! –  The Chaz 2.0 Sep 3 '11 at 20:51

Edit. Turns out this limit does not exist, because the one-sided limits as $x \to 0+$ and $x \to 0-$ are different. See @Américo's answer for more details.

The problem in your idea will become apparent once you take logs: Define $L$ to be the limit. Then $$ \ln L = \lim_{x \to 0} \frac{\ln(\frac{\sin x}{x})}{x^3} . $$

Now, it is true that the numerator approaches $0$ as $x \to 0$. But I cannot immediately conclude that the limit is $0$, since this is a $0/0$ indeterminate form. Can you spot the connection to your question here?

I presume you should be more comfortable in evaluating $0/0$ indeterminate forms. Can you take it from here? As a starting point, you might want to get rid of the $\ln$ by using some standard limit theorems. (Hint: $\ln(1+h)$ as $h$ goes to $0$.)

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Hint: Rewrite $\left( \dfrac{\sin x}{x}\right) ^{1/x^{3}}$ as $$\left( \frac{\sin x}{x}\right) ^{1/x^{3}}=e^{\left( \ln \left( \frac{\sin x}{x} \right) \right) /x^{3}}$$

and look at the side limits: $$\lim_{x\rightarrow 0^{+}}e^{\left( \ln \left( \frac{\sin x}{x}\right) \right) /x^{3}}\neq \lim_{x\rightarrow 0^{-}}e^{\left( \ln \left( \frac{ \sin x}{x}\right) \right) /x^{3}}.$$

Additional hint: show that

$$\lim_{x\rightarrow 0^{+}}\left( \ln \left( \frac{\sin x}{x}\right) \right) /x^{3}=-\infty ,$$

$$\lim_{x\rightarrow 0^{-}}\left( \ln \left( \frac{\sin x}{x}\right) \right) /x^{3}=+\infty .$$

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I suppose $1/x^2$ would be the correct exponent if I want the limit to exist. –  Srivatsan Sep 3 '11 at 15:29
    
@Srivatsan Narayanan: or $1/x^{2n}$. –  Américo Tavares Sep 3 '11 at 15:41
    
@Bill Dubuque: Thanks! That's the 2nd time I leave trash (%). –  Américo Tavares Sep 3 '11 at 16:06

Hint: If allowed, I would use the Maclaurin series for $\sin(x)=x+\frac{x^3}{6}+O(x^5)$. Then, after diving by $x$, note that $\lim\limits_{x\to0}(1+x^2)^{1/x^3}=\lim\limits_{x\to0}(1+x^2)^{(1/x^2)(1/x)}=\lim\limits_{x\to0}e^{1/x}$.

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